(x+1/x)^2-5(x+1/x)+6=0

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Solution for (x+1/x)^2-5(x+1/x)+6=0 equation:


D( x )

x = 0

x = 0

x = 0

x in (-oo:0) U (0:+oo)

(x+1/x)^2-(5*(x+1/x))+6 = 0

(x+1/x)^2-5*(x+1/x)+6 = 0

(x+x^-1)^2-5*(x+x^-1)+6 = 0

(x+x^-1)^2-5*(x+x^-1)+6 = 0

x^2-5*x-5*x^-1+x^-2+2+6 = 0

x^2-5*x-5*x^-1+x^-2+8 = 0

x^2-5*x-5*x^-1+x^-2+8 = 0

1*x^2-5*x^1-5*x^-1+1*x^-2+8*x^0 = 0

(1*x^4-5*x^3+8*x^2-5*x^1+1*x^0)/(x^2) = 0 // * x^4

x^2*(1*x^4-5*x^3+8*x^2-5*x^1+1*x^0) = 0

x^2

x^4-5*x^3+8*x^2-5*x+1 = 0

{ 1, -1 }

1

x = 1

x^4-5*x^3+8*x^2-5*x+1 = 0

1

x-1

x^3-4*x^2+4*x-1

x^4-5*x^3+8*x^2-5*x+1

x-1

x^3-x^4

8*x^2-4*x^3-5*x+1

4*x^3-4*x^2

4*x^2-5*x+1

4*x-4*x^2

1-x

x-1

0

x^3-4*x^2+4*x-1 = 0

{ 1, -1 }

1

x = 1

x^3-4*x^2+4*x-1 = 0

1

x-1

x^2-3*x+1

x^3-4*x^2+4*x-1

x-1

x^2-x^3

4*x-3*x^2-1

3*x^2-3*x

x-1

1-x

0

x^2-3*x+1 = 0

DELTA = (-3)^2-(1*1*4)

DELTA = 5

DELTA > 0

x = (5^(1/2)+3)/(1*2) or x = (3-5^(1/2))/(1*2)

x = (5^(1/2)+3)/2 or x = (3-5^(1/2))/2

x in { (3-5^(1/2))/2, (5^(1/2)+3)/2, 1, 1}

(x-((3-5^(1/2))/2))*(x-((5^(1/2)+3)/2))*(x-1)^2 = 0

(x-((3-5^(1/2))/2))*(x-((5^(1/2)+3)/2))*(x-1)^2 = 0

( x-((3-5^(1/2))/2) )

x-((3-5^(1/2))/2) = 0 // + (3-5^(1/2))/2

x = (3-5^(1/2))/2

( x-((5^(1/2)+3)/2) )

x-((5^(1/2)+3)/2) = 0 // + (5^(1/2)+3)/2

x = (5^(1/2)+3)/2

( x-1 )

x-1 = 0 // + 1

x = 1

x in { (3-5^(1/2))/2, (5^(1/2)+3)/2, 1, 1 }

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